CHEM-120 Final Answers.docx

Week 8 : Final Exam - Final Exam
CHEM120Final-Exam-Page-3




1. Question: (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the
number of grams of H3PO4 needed to prepare 0.25L of a 0.2M H3PO4 solution. Show your work.
(b, 5 pts) What volume, in Liters, of a 0.2 M H3PO4 solution can be prepared by diluting 50 mL of a 5M
H3PO4 solution? Show your work. (Pts. : 10)


A. Molarity = moles of solute/liters of solution
moles of solute = 0.2 M*0.25 L = 0.05 mol H3PO4

Using the molar mass given, convert this amount to grams.
mass = 0.05 mol * (97.994 g/mol) = 4.89 grams H3PO4




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er as
B. C1V1 = C2V2. C1 = 5 M, V1=0.05L, C2 = 0.2M; V2 = [(5M)(0.05L)]/(0.2M) =




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1.25L




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2. Question: (TCO 7) (a, 5 pts) What is the mass/volume percent of a solution prepared by dissolving
43 g of NaOH in enough water to make a final volume of 120 mL? Show your work.
(b, 5 pts) How many mL of a 10% solution can be made from the solution in part a? Show your work.
2. First convert the given mass of NaOH to volume (in mL) using the density of
ed d




NaOH w hich is 2.13 g/mL.
Volume = 43 g * (1 mL/2.13 g) = 20.19 mL
ar stu




Volume % = (volume of solute / volume of solution) * 100%
Volume % = (20.19 mL/120 mL) * 100% = 16.8 %
is




b. Volume % = volume of NaOH/ total volume
0.10 = 20.19 mL/total volume
Th




Solving for total volume yields:

V_total = 201.88 mL
So, about 202 mL of a 10% solution can be made from the solution in part a.
sh




(Pts. : 10)




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